1. Reference
These slides, with minor modification and some deletion, come from U. of Delaware – and the web, of course.
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2. Static Single Assignment
SSA: A program is in SSA form iff
Each variable is statically defined exactly only once, and
Each use of a variable is dominated by that variable’s definition.
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3. Example x  X  1 X  2 X3  (X1, X2) 4
In general, how to transform an arbitrary program into SSA form?
Does the definition of X2 dominates its use in the example?
X  2
x 
X3  (X1, X2) 4
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4. SSA: Motivation
Provide a uniform basis of an IR to solve a wide range of classical dataflow problems
Encode both dataflow and control flow information
A SSA form can be constructed and maintained efficiently
Many SSA dataflow analysis algorithms are more efficient (have lower complexity) than their CFG counterparts.
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5. Static Single-Assignment Form
Each variable has only one definition in the program text.
This single static definition can be in a loop and
may be executed many times. Thus, even in a
program expressed in SSA, a variable can be
dynamically defined many times.
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6. Advantages of SSA Simpler dataflow analysis
No need to use use-def/def-use chains, which requires NM space for N uses and M definitions
SSA form relates in a useful way with dominance structures.
Everybody does it !
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7. SSA Form – An Example ? SSA-form Each name is defined exactly once
Each use refers to exactly one name
What’s hard
Straight-line code is trivial
Splits in the CFG are trivial
Joins in the CFG are hard
Building SSA Form
Insert Ø-functions at birth points
Rename all values for uniqueness
x 
x  a + b
x  y - z
x  13
z  x * q ?
s  w - x
[Courtesy: Slide are from the book wibesite
from Prof. K. Cooper’s website]
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8. Birth Points Consider the flow of values in this example:
The value x appears everywhere
It takes on several values.
Here, x can be 13, y-z, or 17-4
Here, it can also be a+b
If each value has its own name …
Need a way to merge these
distinct values
Values are “born” at merge points
x  a + b
x  y - z
x  13
z  x * q
s  w - x
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9. Birth Points Consider the flow of values in this example:
New value for x here
or y - z
x  a + b
x  y - z
x  13
New value for x here
13 or ( or y - z)
z  x * q
New value for x here
a+b or ((13 or (17-4 or y-z))
s  w - x
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10. Consider the value flow below:
Birth Points (cont.)
Consider the value flow below:
x 
x  a + b
x  y - z
These are all birth points for values
All birth points are join points
Not all join points are birth points
Birth points are value-specific …
x  13
z  x * q
s  w - x
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11. Review SSA-form Each name is defined exactly once
A Ø-function is a special kind of copy that selects one of its parameters.
The choice of parameter is governed by the CFG edge along which control reached the current block.
Real machines do not implement a Ø-function directly in hardware.(not yet!)
SSA-form
Each name is defined exactly once
Each use refers to exactly one name
What’s hard
Straight-line code is trivial
Splits in the CFG are trivial
Joins in the CFG are hard
Building SSA Form
Insert Ø-functions at birth points
Rename all values for uniqueness
y1  ...
y2  ...
y3  Ø(y1,y2)
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12. SSA Form in Control-Flow Path Merges
Is this code in SSA form? B1
b  M[x]
a  0
No, two definitions of a appear in
the code (in B1 and B3) B2
if b<4
How can we transform this code
into a code in SSA form? B3
a  b
We can create two versions of
a, one for B1 and another for B3. B4
c  a + b
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13. SSA Form in Control-Flow Path Merges
But which version should we
use in B4 now? B1
b  M[x]
a1  0
We define a fictional function that
“knows” which control path was
taken to reach the basic block B4: B2
if b<4 B3
a2  b B4
c  a? + b a1 if we
arrive at B4
from B2 f
( )
a1, a2 = a2 if we
arrive at B4
from B3
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14. SSA Form in Control-Flow Path Merges
But, which version should we
use in B4 now? B1
b  M[x]
a1  0
We define a fictional function that
“knows” which control path was
taken to reach the basic block B4: B2
if b<4 B3
a2  b B4
a3  (a2,a1)
c  a3 + b a1 if we
arrive at B4
from B2 f ( )
a2, a1 = a2 if we
arrive at B4
from B3
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15. A Loop Example a  0 a1  0 a3  (a1,a2) b2  (b0,b2) c2  (c0,c1)
b2  a3+1
c1  c2+b2
a2  b2*2
if a2 < N
return
b  a+1
c  c+b
a  b*2
if a < N
return
(b0,b2) is not necessary because b0 is
never used. But the phase that generates
 functions does not know it.
Unnecessary functions
are eliminated by dead code elimination.
Note: only a,c are first used in
the loop body before it is redefined.
For b, it is redefined right at the
Beginning!
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16. The  function How can we implement a  function that “knows”
which control path was taken?
Answer 1: We don’t!! The  function is used only
to connect use to definitions during
optimization, but is never implemented.
Answer 2: If we must execute the  function, we can
implement it by inserting MOVE instructions
in all control paths.
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17. Criteria For Inserting  Functions
We could insert one  function for each variable
at every join point(a point in the CFG with more
than one predecessor). But that would be wasteful.
What should be our criteria to insert a  function
for a variable a at node z of the CFG?
Intuitively, we should add a function  if
there are two definitions of a that can reach
the point z through distinct paths.
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18. Path Convergence Criterion
Insert a  function for a variable a at a node z if
all the following conditions are true:
1. There is a block x that defines a
2. There is a block y  x that defines a
3. There is a non-empty path Pxz from x to z
4. There is a non-empty path Pyx from y to z
5. Paths Pxz and Pyz don’t have any nodes in
common other than z
6. The node z does not appear within both Pxz
and Pyz prior to the end, but it might appear
in one or the other.
The start node contains an implicit definition
of every variable.
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19. Iterated Path-Convergence Criterion
The  function itself is a definition of a.
Therefore the path-convergence criterion
is a set of equations that must be satisfied.
while there are nodes x, y, z satisfying conditions 1-5
and z does not contain a  function for a
do insert a (a, a, …, a) at node z
This algorithm is extremely costly, because it
requires the examination of every triple of
nodes x, y, z and every path leading from
x to y.
Can we do better? – a topic for more discussion
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